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在組織標籤之間列印文字

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在組織標籤之間列印文字

我想編寫一個 bash 函數，列印文件中匹配的行## mode: org和之間包含的文字部分，各部分之間有一個空白行。## # End of org在 之前##，可以有任意數量的空格。

以下是從中提取資訊的文件範例。

file: test.sh

## mode: org
## * Using case statement
## # End of org
case $arg in
 ("V")
   echo "Author"
   ;;
 (*)
   ## mode: org
   ## ** Silent Error Reporting Mode (SERM) in getopts
   ## *** Detects warnings without printing built-in messages.
   ## *** Enabled by colon {:} as first character in shortopts.
   ## # End of org
   break
   ;;
esac

期望的輸出是

* Using case statement

** Silent Error Reporting Mode (SERM) in getopts
*** Detects warnings without printing built-in messages.
*** Enabled by colon {:} as first character in shortopts.

我已經做到了

capture-org ()
{
  sed -n '/^ *## mode: org$/,/^ *## # End of org$/s/ *//p' "$1" |
   sed 's/^## mode: org$/\n## mode: org/' |
   sed '/^## mode: org$/d' | sed '/^## # End of org$/d' | cut -c 3-
}

能以更簡潔的方式完成嗎？

答案1

sed '/^[[:space:]]*## mode: org$/,/^[[:space:]]*## # End of org$/!d; /^[[:space:]]*## mode: org$/d; s/^[[:space:]]*## # End of org$//g; s/^[[:space:]]*## //'

這將刪除開始和結束模式之外的所有內容。然後它將刪除開始模式並用空白行替換結束模式（成為區塊分隔符號）。

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